Two crates, one with mass m1=6 kg and the other with mass m2=10 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (as shown in Figure). A woman wearing golf shoes (for traction) pulls horizontally on the 10 kg crate with a force F that gives the crate an acceleration of 1.5 m/s2.
(a) What is the acceleration( in m/s2) of the 6 kg crate?
(b) Find the tension T (in N) in the rope that connects the two crates.
(c) Calculate the magnitude of the force F (in N).
(a) same as 10 kg crate
(b) T = 6 kg * 1.5 m/s^2 = ? N
(c) F = (10 kg + 6 kg) * 1.5 m/s^2
To solve this problem, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass.
(a) The acceleration of the 6 kg crate can be determined using the same force applied to the 10 kg crate. Since the crates are connected by a rope, they will accelerate together. Therefore, their acceleration must be the same. So, the acceleration of the 6 kg crate is also 1.5 m/s^2.
(b) To find the tension T in the rope, we need to consider the forces acting on the 6 kg crate. The only horizontal force acting on it is the tension in the rope. According to Newton's second law, we can write:
T - ma = 0
where T is the tension, m is the mass of the 6 kg crate, and a is its acceleration.
Substituting the values, T - 6 kg * 1.5 m/s^2 = 0, we can solve for T:
T = 6 kg * 1.5 m/s^2 = 9 N
Therefore, the tension in the rope connecting the two crates is 9 N.
(c) The magnitude of force F can be determined using the 10 kg crate's mass and acceleration. Again, using Newton's second law:
F = ma = 10 kg * 1.5 m/s^2 = 15 N
So, the magnitude of force F is 15 N.
(a) To find the acceleration of the 6 kg crate, we can use the concept of Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Since there are no external horizontal forces acting on the system of crates, the tension in the rope is the only force acting horizontally. It is also the force responsible for accelerating both crates. Therefore, the acceleration of the 6 kg crate will be the same as the acceleration of the 10 kg crate.
So, the acceleration of the 6 kg crate is 1.5 m/s^2.
(b) To find the tension T in the rope, we can consider the forces acting on the 10 kg crate. The force F applied by the woman and the tension T in the rope are the only horizontal forces acting on the crate.
Using Newton's second law for the 10 kg crate, we have:
F - T = m2 * a
Substituting the given values, we get:
F - T = 10 kg * 1.5 m/s^2
Simplifying the equation, we have:
F - T = 15 N
Since the tension in the rope is equal to the tension in the opposite direction, we can write:
T = F - 15 N
(c) To calculate the magnitude of force F applied by the woman, we can sum up the forces acting on the 10 kg crate:
F - T = m2 * a
Substituting the known values, we have:
F - T = 10 kg * 1.5 m/s^2
Substituting T = F - 15 N (from part b), we can rewrite the equation as:
F - (F - 15 N) = 10 kg * 1.5 m/s^2
Simplifying the equation, we have:
15 N = 15 N
This means that the magnitude of force F applied by the woman is 15 N.